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Eğitmen, öğrencilere sınav hazırlığı için çeşitli matematik konularını adım adım anlatmaktadır.","Video, benzerlik, dik üçgenler, Pisagor bağıntısı, Öklit bağıntısı ve algoritma konularını içermektedir. Eğitmen, her konuyu örnek sorular üzerinden açıklamakta ve çözüm yöntemlerini detaylı olarak göstermektedir. Video, iki farklı sınav senaryosu üzerinden hazırlanmış olup, her senaryoda farklı matematik problemleri çözülmektedir.","Videoda ayrıca karelerin özellikleri, eş üçgenler, gölgeler, paralel kenarlar, özel üçgenler (45-45-90 ve 30-60-90) ve sınav puan hesaplama algoritması gibi konular ele alınmaktadır. Eğitmen, öğrencilerin sınavda başarılı olmaları için motivasyon vererek, her problemi adım adım çözmekte ve çözüm stratejilerini paylaşmaktadır."]},"endTime":2476,"title":"Matematik Soru Çözüm Dersi","beginTime":0}],"fullResult":[{"index":0,"title":"Yazılı Hazırlık ve Video İçeriği","list":{"type":"unordered","items":["Yazılıdan çıkan soruların Selim Hoca'nın çözdüğü soruların çok benzerleri olduğu belirtiliyor.","İki farklı senaryo hazırlanmış ve ilk senaryo çözülecek.","Kanalın ana sayfasında tekrar videosu yayınlanmış ve on tane sınavda karşınıza gelebilecek soru çözülecek."]},"beginTime":0,"endTime":48,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=0&ask_summarization=1"},{"index":1,"title":"PDF ve Video Detayları","list":{"type":"unordered","items":["Açıklamalara PDF dosyası konulmuş ve sınavda karşınıza gelebilecek her şeyin bu PDF'de olduğu belirtiliyor.","On tane soru hazırlanmış ve birlikte çözülecek.","İzleyicilerden beğeni butonuna basmaları, abone olmaları ve sonraki senaryoyu izlemeleri isteniyor."]},"beginTime":48,"endTime":82,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=48&ask_summarization=1"},{"index":2,"title":"İlk Soru Çözümü","list":{"type":"unordered","items":["İlk soruda A, B, C, D ve EFGH birer kare olduğu belirtiliyor.","Karenin bütün kenarlarının birbirine eşit olduğu vurgulanıyor.","EFGH karesinin köşelerinin A, B, C, D karesinin kenarları üzerinde olduğu ve A, B, C, D karesinin çevresinin bulunması isteniyor."]},"beginTime":82,"endTime":112,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=82&ask_summarization=1"},{"index":3,"title":"Karenin Özellikleri","list":{"type":"unordered","items":["Karenin en önemli özelliği bütün açılarının 90 derece olmasıdır.","Kare deyince hemen bütün kenarlarının eşit olduğunu işaretlemek gerekir."]},"beginTime":114,"endTime":134,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=114&ask_summarization=1"},{"index":4,"title":"Eşlik ve Benzerlik Kavramları","list":{"type":"unordered","items":["Eşlik ve benzerlik aslında aynı şeydir.","Benzerlik oranı bir ise üçgenlere eş denir."]},"beginTime":134,"endTime":145,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=134&ask_summarization=1"},{"index":5,"title":"Üçgenlerin Açıları ve Benzerliği","list":{"type":"unordered","items":["Bir açısı 90 derece olan üçgenin diğer iki açısının toplamı da 90 derecedir.","Karede açılar zincirleme olarak 90 derece olur.","Tüm üçgenler benzerdir çünkü açıları birbirine eşittir."]},"beginTime":145,"endTime":224,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=145&ask_summarization=1"},{"index":6,"title":"Eş Üçgenlerin Özellikleri","list":{"type":"unordered","items":["90 derecenin karşısında karenin bir kenarı vardır.","Açıları birbirine eşit olduğu için 90 derecenin karşıları da aynı olduğu için dört kenar birbirine eşittir.","Benzerlik oranı bir olduğundan, tüm üçgenlerde 90 derecenin karşısındaki kenarlar eşittir."]},"beginTime":224,"endTime":288,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=224&ask_summarization=1"},{"index":7,"title":"Karenin Çevresinin Hesaplanması","list":{"type":"unordered","items":["Büyük A, B, C, D karesinin çevresi sorulmaktadır.","Bir kenarının 4 olduğunu bulduk çünkü 8+6=14'tür.","Dört tane 4 olduğu için çevresi 14×4=56 olarak hesaplanmıştır."]},"beginTime":288,"endTime":304,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=288&ask_summarization=1"},{"index":8,"title":"Merdiven Problemi Çözümü","list":{"type":"unordered","items":["Zemine dik konumda duvara dayalı 50 metre uzunluğunda bir merdiven, temas ettiği A noktasının yerden yüksekliği 30 metredir.","Merdiven A noktasından aşağı doğru 16 metre kayarak yeni bir pozisyona gelmiştir.","Merdivenin uzunluğu değişmediği için, 50 metre uzunluğunda olan merdivenin yeni durumunda B ile B' arası uzaklık 8 metredir."]},"beginTime":313,"endTime":471,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=313&ask_summarization=1"},{"index":9,"title":"Park Lambası ve Ağaçlar Problemi","list":{"type":"unordered","items":["Bir parkta doğrusal olarak sıralanmış 18 metre yüksekliğinde bir lamba, 3 metre ve 6 metre yüksekliğinde iki ağaç bulunmaktadır.","Gece lamba yanarken 6 metrelik ağacın gölge boyu, 3 metrelik ağacın gölge boyunun iki katı olmaktadır.","Ağaçlar arasındaki uzaklık 27 metredir ve soruda ağaç gölgelerinin uzunlukları toplamı istenmektedir."]},"beginTime":471,"endTime":612,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=471&ask_summarization=1"},{"index":10,"title":"Benzer Üçgenler Problemi Çözümü","list":{"type":"unordered","items":["İki benzer üçgen arasında 3/18 oranı bulunuyor ve bu oran kullanılarak a/a+b oranı hesaplanıyor.","a/a+b oranı 1/6 olarak sadeleştirildiğinde, aşağıdaki uzunluğun yukarıdaki uzunluğun 6 katı olduğu anlaşılıyor.","Benzerlik oranı kullanılarak b=5a ve c=4a bulunuyor, toplam 9a=27 denkleminden a=3 olarak hesaplanıyor."]},"beginTime":616,"endTime":755,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=616&ask_summarization=1"},{"index":11,"title":"Gölgelerin Boylarının Toplamı","list":{"type":"unordered","items":["Sorunun cevabı gölgelerin boylarının toplamı olarak 3a olarak bulunuyor ve a=3 olduğundan toplam 9 olarak hesaplanıyor.","Şekli çizmek zorunlu değil, doğrudan benzerlik oranları kullanılarak çözüm yapılabilir.","Benzer üçgenler arasındaki oranlar kullanılarak a=3 değeri bulunuyor ve gölgelerin boyları toplamı 9 olarak hesaplanıyor."]},"beginTime":755,"endTime":885,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=755&ask_summarization=1"},{"index":12,"title":"Paralel Doğrular ve Benzerlik","list":{"type":"unordered","items":["Paralel doğrular ve benzerlik konusu önemli bir konu olarak vurgulanıyor.","FC'nin üç katının DF'ye eşit olduğu bilgisi veriliyor, yani FC=a ise DF=3a olarak hesaplanıyor.","Benzerlik oranları kullanılarak 12/4=3 ilişkisi bulunuyor."]},"beginTime":885,"endTime":932,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=885&ask_summarization=1"},{"index":13,"title":"Benzerlik Problemi Çözümü","list":{"type":"unordered","items":["Pales çıktığında kenarlardan birine paralel atılması gerekiyor.","Üçgende benzerlik oranı 3a/4a olarak hesaplanıyor ve bu oran 12/(12+y) ile eşitleniyor.","Benzerlik oranı kullanılarak y=4 bulunuyor ve n=6 olarak hesaplanıyor."]},"beginTime":932,"endTime":1086,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=932&ask_summarization=1"},{"index":14,"title":"İkinci Benzerlik Problemi","list":{"type":"unordered","items":["Paralel çizgiler sayesinde üçgenlerde benzerlik oluşuyor ve benzerlik oranı 5/7 olarak hesaplanıyor.","Benzerlik oranı kullanılarak kenar uzunlukları 5a ve 7a olarak ifade ediliyor.","İkinci üçgende benzerlik oranı kullanılarak x=21 olarak bulunuyor."]},"beginTime":1086,"endTime":1184,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1086&ask_summarization=1"},{"index":15,"title":"Pisagor Bağıntısı Sorusu","list":{"type":"unordered","items":["Dik üçgende Pisagor bağıntısı kullanılarak hipotenüs hesaplanıyor: a² = √82² + 3² = 82 + 9 = 91, a = √91.","İkinci dik üçgende Pisagor bağıntısı uygulanarak b² = a² + 3² = 91 + 9 = 100, b = 10.","Son dik üçgende Pisagor bağıntısı kullanılarak x² = b² + 7² = 100 + 49 = 51, x = √51."]},"beginTime":1190,"endTime":1310,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1190&ask_summarization=1"},{"index":16,"title":"Öklid Bağıntısı","list":{"type":"unordered","items":["Öklid bağıntısı için üçgenin dik üçgen olması ve dik açıdan dik indirilmesi gerekiyor.","Ana bağıntı: Dikten indirilen yüksekliğin karesi, hipotenüsteki ayırdığı parçaların çarpımına eşittir (h² = p × k).","Diğer bağıntılar: Dik kolun karesi = kendi bölümündeki parça × tamamı (a² = p × (p+k)), yan kolun karesi = kendi bölümündeki parça × tamamı (b² = k × (p+a))."]},"beginTime":1310,"endTime":1394,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1310&ask_summarization=1"},{"index":17,"title":"Dik Üçgen Problemi Çözümü","list":{"type":"unordered","items":["Dik üçgende dikten dikilmiş bir inmiş ve yüksekliğin karesi, ayırdığı parçaların çarpımına eşittir.","Küçük üçgende yüksekliğin karesi 16 olarak hesaplanarak h'nin 4 olduğu bulunmuştur.","Büyük üçgende yükseklik 8 olarak belirlenmiş ve 8² = (x+2)×8 denklemi çözülerek x'in 6 olduğu hesaplanmıştır."]},"beginTime":1396,"endTime":1495,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1396&ask_summarization=1"},{"index":18,"title":"İkizkenar Dik Üçgen Özellikleri","list":{"type":"unordered","items":["8-8-8 üçgeni ikizkenar dik üçgen olarak belirlenmiştir.","İkizkenar dik üçgende hipotenüs, dik kolların kök 2 katıdır.","Y'nin karesi, kendi bölümündeki parça çarpı tamamı formülüyle hesaplanarak 128 bulunmuştur."]},"beginTime":1495,"endTime":1578,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1495&ask_summarization=1"},{"index":19,"title":"Sonuç ve Algoritma Sorusu","list":{"type":"unordered","items":["45-45-90 üçgeninde hipotenüs, dik kolların kök 2 katıdır ve bu özellik kullanılarak çözüm tamamlanmıştır.","Algoritma sorusu büyük ihtimalle karşınıza gelecektir."]},"beginTime":1578,"endTime":1594,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1578&ask_summarization=1"},{"index":20,"title":"Sınav Puanlama Algoritması","list":{"type":"unordered","items":["Yüz soruluk bir sınavın sonunda belirli ölçütlere göre puan hesaplayan ve dönüt veren bir algoritma işleyişi verilmiştir.","Sınav sonucunda doğru sayısı, yanlış sayısı ve boş sayısı girilerek net sayısı hesaplanır.","Net sayısı, doğru sayısından yanlışların dörde bölünmesi çıkarılarak 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389'dur."]},"beginTime":1837,"endTime":1950,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1837&ask_summarization=1"},{"index":24,"title":"Benzerlik Sorusu Çözümü","list":{"type":"unordered","items":["ABC üçgeni şeklinde modellenmiş bir dağın B ve C noktalarından zirveye doğru tırmanmaya başlanmıştır.","C noktasından harekete başlayan dağcı yolun 4/5'ini tırmanmış ve D noktasında 1600 metre yüksekliğe ulaşmıştır.","B noktasından hareket eden dağcı yolun yarısında bulunan E'ye gelmiş ve dağın zirvesinin yüksekliği 2000 metre olarak hesaplanmıştır."]},"beginTime":1950,"endTime":2121,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1950&ask_summarization=1"},{"index":25,"title":"Üçgen Benzerliği Problemi","list":{"type":"unordered","items":["ABC üçgeninin kenarları üzerinde DHGF noktaları işaretlenmiş ve bazı uzunluklar verilmiştir: AD=DB=7/2, AF=2, GH=4, BC=15, EC=3.","Soruda DFGH noktalarından herhangi ikisini birleştirerek ABC üçgenine benzer üçgen elde edilmesi ve benzerlik oranının açıklanması istenmektedir.","İki farklı benzerlik durumu bulunabilir ve her iki durumda da benzerlik oranı 1/6 olarak hesaplanmıştır."]},"beginTime":2125,"endTime":2254,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=2125&ask_summarization=1"},{"index":26,"title":"İlk Benzerlik Durumu","list":{"type":"unordered","items":["İlk benzerlik durumunda D ile G noktaları birleştirilerek orta taban oluşturulmuştur.","Benzerlik oranı 6/2=1/2 olarak hesaplanmıştır.","ABC üçgeni ile ADG üçgeni kenar-açı-kenar (KAK) aksiyonuna göre benzerdir."]},"beginTime":2254,"endTime":2345,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=2254&ask_summarization=1"},{"index":27,"title":"İkinci Benzerlik Durumu","list":{"type":"unordered","items":["İkinci benzerlik durumunda 2 ile 15 arasındaki 5 kat oran kullanılmıştır.","Benzerlik oranı 2/12 veya 3/18 olarak hesaplanmıştır.","ABC üçgeni ile CAF üçgeni kenar-açı-kenar (KAK) aksiyonuna göre benzerdir."]},"beginTime":2345,"endTime":2470,"href":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=2345&ask_summarization=1"}],"linkTemplate":"/video/preview/13045130009727306631?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"9th grade mathematics 2nd term 1st written 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Öğretmen, 11. sınıf öğrencilerine üniversite sınavına hazırlık amacıyla yazılı sınavı çözmektedir.","Video, parabol konusundan başlayarak eşitsizlikler konusuna kadar uzanan kapsamlı bir matematik dersidir. İlk bölümde parabolün en büyük ve en küçük değerleri, tepe noktası bulma formülü ve parabol-doğru ilişkileri ele alınırken, sonraki bölümlerde denklem sistemleri ve eşitsizliklerin çözüm yöntemleri adım adım gösterilmektedir.","Videoda her konu için farklı çözüm yöntemleri sunulmakta, parabol fonksiyonlarının grafik çizimi, denklem sistemlerinin yerine koyma metodu ile çözümü ve eşitsizliklerin çarpanlara ayırma, kök bulma ve işaret tablosu oluşturma yöntemleriyle çözümü detaylı olarak anlatılmaktadır. Video, üniversite sınavına hazırlık kapsamında hazırlanmış olup, öğrencilere sınav başarıları dilekleriyle sonlanmaktadır."]},"endTime":2501,"title":"11. Sınıf Matematik Yazılı Sınavı Hazırlık Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"11. Sınıf Matematik Yazılı Senaryosu","list":{"type":"unordered","items":["11. sınıf öğrencileri için üniversite sınavına hazırlanmadan önce short otorite başarı puanlarını yükseltmek amacıyla bir yazılı senaryosu çözülecek.","Senaryoda dokuz soru yer alacak ve bunlardan üçü parabol konusundan olacak.","Parabol konusundan soru gelecekse, en büyük-en küçük değer ile ilgili üç farklı durum işlenecek."]},"beginTime":0,"endTime":81,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=0&ask_summarization=1"},{"index":1,"title":"Parabolün En Büyük Değeri Sorusu","list":{"type":"unordered","items":["Bir balonun çıkabileceği maksimum yüksekliği hesaplamak için verilen bağıntıda r (tepe noktasının apsisi) -b/2a formülüyle bulunuyor.","r = -24/(-6) = 4 olarak hesaplanıyor ve bu değer fonksiyonda yerine konularak k (tepe noktasının ordinatı) bulunuyor.","h(4) = -3(4)² + 4(4) + 160 = 208 metre olarak hesaplanıyor, bu da balonun yerden en fazla çıkabileceği yükseklik."]},"beginTime":81,"endTime":202,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=81&ask_summarization=1"},{"index":2,"title":"Dik Üçgenin Alanı Sorusu","list":{"type":"unordered","items":["Dik kenar uzunlukları 10x ve 2x+4 olan dik üçgenin alanı (2x+4)(10-x)/2 formülüyle hesaplanıyor.","Alan fonksiyonu -x² + 8x + 20 olarak bulunuyor ve kolları aşağı doğru bir parabol olduğu için en büyük değeri tepe noktasında bulunuyor.","r = -8/(-2) = 4 olarak hesaplanıyor ve h(4) = -16 + 32 + 20 = 36 metre olarak bulunuyor, bu da üçgenin en büyük alanı."]},"beginTime":202,"endTime":394,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=202&ask_summarization=1"},{"index":3,"title":"Kar Hesaplama Sorusu","list":{"type":"unordered","items":["Bir ürünün alış fiyatı x² - 2x + 3, satış fiyatı 2x² + 6x + 30 olarak verilmiş ve en az kar miktarı soruluyor.","Kar fonksiyonu k(x) = 2x² + 6x + 30 - (x² - 2x + 3) = x² + 8x + 27 olarak bulunuyor.","Kolları yukarı doğru bir parabol olduğu için en küçük değeri tepe noktasında bulunuyor, r = -8/2 = -4 ve k(-4) = 11 olarak hesaplanıyor."]},"beginTime":394,"endTime":548,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=394&ask_summarization=1"},{"index":4,"title":"Parabol ve Doğru Arasındaki İlişki","list":{"type":"unordered","items":["Parabol ile doğru arasındaki üç durum vardır: iki farklı noktada kesişme, teğet olma ve kesişmeme.","Parabol ve doğrunun denklemleri birbirine eşitlenerek ortak çözüm denklemi oluşturulur.","Ortak çözüm denkleminin deltası sıfırdan büyükse iki farklı noktada kesişir, sıfıra eşitse teğet olur, sıfırdan küçükse kesişmez."]},"beginTime":550,"endTime":636,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=550&ask_summarization=1"},{"index":5,"title":"Teğet Olma Koşulu","list":{"type":"unordered","items":["Teğet olma durumunda ortak çözüm denkleminin deltası sıfıra eşit olmalıdır.","Delta hesaplaması için b² - 4ac formülü kullanılır.","Verilen örnekte a = 3/2 olduğunda parabol ve doğru teğet olur."]},"beginTime":636,"endTime":729,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=636&ask_summarization=1"},{"index":6,"title":"Parabol ve Öteleme Dönüşümleri","list":{"type":"unordered","items":["Parabolün tepe noktasını bulmak için x ekseninin ortasını alırız.","Parabole dışarıdan sayı eklenirse, tepe noktası o sayı kadar yukarı veya aşağı öteleşir.","Parabolün denklemini bulmak için baş katsayısı, tepe noktası ve eksenleri kestiği noktalar kullanılır."]},"beginTime":729,"endTime":1087,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=729&ask_summarization=1"},{"index":7,"title":"Parabol Denkleminin Farklı Yöntemleri","list":{"type":"unordered","items":["Parabolün denklemini yazarken köklerini biliyorsak, f(x) = a(x-x₁)(x-x₂) formülü kullanılır.","Parabolün tam kare hali f(x) = a(x-r)² + k şeklinde yazılabilir.","Parabolün denklemini bulduktan sonra tepe noktası ve eksenleri kestiği noktalar hesaplanabilir."]},"beginTime":1087,"endTime":1203,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1087&ask_summarization=1"},{"index":8,"title":"Parabol Denkleminin Çözümü","list":{"type":"unordered","items":["Parabol denklemi x = -((x-1)²+9) şeklinde verilmiş ve x yerine x-2 yazarak f(x-2) = -((x-3)²+4) elde edilmiştir.","Parabolün tepe noktası (3,4) olarak bulunmuş ve y eksenini -5 noktasında kesmiştir.","Parabol x eksenini 1 ve 5 noktalarında kesmiştir."]},"beginTime":1208,"endTime":1368,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1208&ask_summarization=1"},{"index":9,"title":"Parabol Denkleminin Alternatif Çözümü","list":{"type":"unordered","items":["Parabolün tepe noktası (1,9) olarak verilmiş ve x yerine x-2 yazarak 2 birim sağa ötelenmiştir.","Eksi 5 değeri uygulandığında 5 birim aşağı ötelenmiştir ve tepe noktası (3,4) olmuştur.","Parabolün denklemi y = -x² + 2x + 4 olarak bulunmuştur."]},"beginTime":1368,"endTime":1452,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1368&ask_summarization=1"},{"index":10,"title":"Denklem Sistemi Çözümü","list":{"type":"unordered","items":["Denklem sistemi x·y = -3 ve x + y = 2 şeklinde verilmiş, yerine koyma metodu kullanılarak çözülmüştür.","x = 3 ve y = -1, x = -1 ve y = 3 çözümleri bulunmuştur.","İkinci denklem sisteminde yerine koyma metodu kullanılarak x = -5/2 ve y = -7/2, x = 3 ve y = 7 çözümleri elde edilmiştir."]},"beginTime":1452,"endTime":1760,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1452&ask_summarization=1"},{"index":11,"title":"Eşitsizlik Çözümü","list":{"type":"unordered","items":["Eşitsizliklerin her durumundan örnekler verilecek ve sınavda karşınıza gelebilecek sorular çözülecek.","İlk eşitsizlik sorusunda x²-3x-18/x²-4x+4 > 0 eşitsizliği çözülüyor.","Eşitsizlik çarpanlarına ayrılarak (x-6)×(x+3)/(x-2)² > 0 şeklinde yazılır ve kökler x=6, x=-3 ve x=2 olarak 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fonksiyonun kökleri bulunur.","f(x) fonksiyonunun kökleri -4 ve 1, paydanın kökleri -2 ve 2 olarak belirlenir.","f(x) fonksiyonunun grafiğine bakarak işaret tablosu oluşturulur ve en büyük dereceli terimin işareti (+) belirlenir."]},"beginTime":1927,"endTime":2129,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=1927&ask_summarization=1"},{"index":14,"title":"Eşitsizlik Çözümü ve Sonuç","list":{"type":"unordered","items":["İşaret tablosu oluşturulduktan sonra f(x)/(x²-4x+4) > 0 eşitsizliğinin çözüm kümesi (-4, -2) ∪ (2, +∞) olarak bulunur.","Eşitsizliği sağlayan en küçük iki tam sayı a ve b için a×b = (-4)×(-3) = 12 olarak hesaplanır.","İkinci eşitsizlik sorusunda (x-7)³/(x-6)×(x-2) ≥ 0 denkleminin çözümü için kökler x=4, x=2 ve x=1 olarak bulunur."]},"beginTime":2129,"endTime":2249,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=2129&ask_summarization=1"},{"index":15,"title":"Eşitsizlik Çözümü ve Sonuç","list":{"type":"unordered","items":["Eşitsizliğin çözüm kümesi (-∞, 1) ∪ [2, 4] olarak bulunur.","Eşitsizliği sağlayan x tam sayı değerlerinin toplamı 1+2+3+4 = 9 olarak hesaplanır."]},"beginTime":2249,"endTime":2318,"href":"/video/preview/7670880032910592829?parent-reqid=1776730733185131-12806496143956599633-balancer-l7leveler-kubr-yp-sas-203-BAL&text=matt2&t=2249&ask_summarization=1"},{"index":16,"title":"Eşitsizlik Sisteminin Çözümü","list":{"type":"unordered","items":["Eşitsizlik sisteminin çözüm kümesi bulunuyor, grafikli olanların aynısı sadece iki denklem var.","İlk eşitsizlik (x(x+6) ≤ 0, x=4 ve x=-6 köklerini buluyor).","İkinci eşitsizlik (5-x)(5+x) > 0, x=5 ve x=-5 köklerini 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